An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?

Given:

Mass of ethylene glycol (C2H6O2) = 222.6 g


Mass of water = 200 g


Density, d = 1.072 g/ml


To find: Molality and Molarity of solution


Formula:


Molality


Molarity, Mo


Density, d


Solution:


Calculation of Molality:


⇒ Molecular Mass of ethylene glycol (C2H6O2) = [12 × 2] + [6 × 1] + [16 × 2]


= 24 + 6 + 32


= 62 g


⇒ Number of moles of ethylene glycol (C2H6O2) = [222.6/62]


= 3.59 moles


⇒ Mass of water = 200 g


⇒ Molality



= 17.95 m


Calculation of Molarity:


⇒ Total mass of solution = [222.6 + 200]


= 422.6 g


From density formula we can find out the volume required.


⇒ Volume of solution, V = [422.6 /1.072]


= 394.216 ml


⇒ Molarity, Mo


⇒ Molarity, Mo


⇒ Molarity = 9.1067


⇒ Molarity ≈ 9.11 M


Therefore the Molality and Molarity of the solution is as follows:


Molality = 17.95 m


Molarity = 9.11 M


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